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Math Help - divisor

  1. #1
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    divisor

    Are we able to say, how do all the divisors of  x^m - 1 like???
    Thanks, nothing on my mind seems good now..
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  2. #2
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    Quote Originally Posted by sidi View Post
    Are we able to say, how do all the divisors of  x^m - 1 like???
    Thanks, nothing on my mind seems good now..
    yes. see cyclotomic polynomials.
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  3. #3
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    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) x^d - 1 as the gcd(x^m - 1,x^n - 1) from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
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  4. #4
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    Quote Originally Posted by sidi View Post

    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) x^d - 1 as the gcd(x^m - 1,x^n - 1) from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
    i see! the one that i gave you: \gcd(x^n - 1, x^m -1)=x^{\gcd(n,m)} - 1. here's a proof: let \gcd(n,m)=d. since d \mid n and d \mid m, we have: x^d -1 \mid x^n - 1 and x^d - 1 \mid x^m - 1.

    now suppose f(x) \mid x^n - 1 and f(x) \mid x^m - 1. if we prove that f(x) \mid x^d - 1, then we're done. we know that there exist r,s \in \mathbb{N} such that rn - sm=d. thus:

    x^{rn} - 1 = x^d(x^m)^s - 1 \equiv x^d - 1 \mod f(x), because x^m \equiv 1 \mod f(x). on the other hand x^{rn}-1 \equiv 0 \mod f(x), because f(x) \mid x^n - 1. hence x^d - 1 \equiv 0 \mod f(x). Q.E.D.
    Last edited by NonCommAlg; April 14th 2009 at 05:04 PM.
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  5. #5
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    Quote Originally Posted by sidi View Post
    Thanks, but
    I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) x^d - 1 as the gcd(x^m - 1,x^n - 1) from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind...
    Here is a similar problem and an alternate solution.
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  6. #6
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    Thank you very much!
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